Question

State De Moivre's theorem and prove it for all integer values?

Answer 1

`(cos theta + i sin theta)^n = cosn theta + isinntheta`

Proof by induction

Explanation:

`(cos theta + i sin theta)^n = cosn theta + isinntheta`

Prove by induction:

Basis case: `n = 1`

`=> ( cos theta + isin theta)^1 = cos theta + i sin theta`

Hence basis case holds

Asume true for `n=k`

`(cos theta + i sin theta)^k = cosk theta + isinktheta`

Showing holds for `n = k+1`

`(cos theta + i sin theta)^(k+1) = (cosk theta + isinktheta)(cos theta + isin theta)`

`(cos theta + i sin theta)^(k+1) =`

`= costheta cosk theta + isinthetacosktheta + isinkthetacostheta - sinktheta sintheta`

`= costhetacosktheta - sinkthetasintheta + i ( sinthetacosktheta + sinktheta cos theta )`

`=> = cos ( (k+1) theta ) + i sin ( (k+1) theta)`

Hence it holds true for `k+1`

It holds for basis case `n=1` and `n = k + 1` for all `k` that holds, hence holds for `n =1`, `n =2` , `n=3` etc...

`=> "holds" AA n in ZZ`

Answer 2

`(cos theta+i sin theta)^n = cos(n theta) + isin(n theta) AA n in RR`

Explanation:

De Moivre's Theorem, states that:

`(cos theta+i sin theta)^n = cos(n theta) + isin(n theta) AA n in RR`

We seek to validate this result for `n in NN`, which we can prove using Mathematical Induction:

Induction Proof - Hypothesis

We seek to prove that:

#

`(cos theta+i sin theta)^n = cos(n theta) + isin(n theta) AA n in NN \ \ \ \` ..... [A]

So let us test this assertion, [A], using Mathematical Induction:

Induction Proof - Base case:

Consider, the special case `n=1` in which case we have the trivial result:

`(cos theta+i sin theta)^1 = cos(1 theta) + isin(1 theta)`

So the given result is true when `n=1`.

Induction Proof - General Case

Now, Let us assume that the given result [A] is true when `n=m`, for some `m in NN, m gt 1`, in which case for this particular value of `m` we have:

`(cos theta+i sin theta)^m = cos(m theta) + isin(m theta) \ \ \` ..... [B]

Then, consider the expression:

`E = (cos theta+i sin theta)^(m+1)`

`\ \ \ = (cos theta+i sin theta)^m(cos theta+i sin theta)`

`\ \ \ = (cos(m theta) + isin(m theta))(cos theta+i sin theta) \ \` (using [B])

`\ \ \ = cos(m theta)cos theta + isin(m theta)cos theta + cos(m theta)i sin theta + isin(m theta)i sin theta`

`\ \ \ = {cos(m theta)cos theta -sin(m theta) sin theta} + i{sin(m theta)cos theta + icos(m theta) sin theta }`

`\ \ \ = cos(m+1) theta + isin(m+1) theta`

Which is the given result [A] with `n=m+1`

Induction Proof - Summary

So, we have shown that if the given result [A] is true for `n=m`, then it is also true for `n=m+1`. But we initially showed that the given result was true for `n=1` so it must also be true for `n=2, n=3, n=4, ...` and so on.

Induction Proof - Conclusion

Then, by the process of mathematical induction the given result [A] is true for `n in NN`

Hence we have:

`(cos theta+i sin theta)^n = cos(n theta) + isin(n theta) AA n in NN \ \ \ \` QED

Answer 3

See below.

Explanation:

First let us prove `r[costheta+jsintheta]`=`re^[jtheta]`.

Let `z=costheta+jsintheta...........`[1]#`, therefore,`dz/[d theta`=`-[sintheta-jcostheta]#

=`j^2sintheta+jcostheta` = `j[costheta+jsintheta]`[ since`j^2=-1]`

But `j[cos theta+jsin theta]`=`z` [ from .....`[1]`, so `dz/[d theta`=`jz`.........`[2]`
Inverting both sides of .....`[2]`, `d theta/[dz`=`1/[jz` and integrating w.r.t `z`

`theta=1/jint1/zdz` = `1/jlnz+c`, but from .....`[1]` when `theta =0,cos theta=1,sin theta=0, i.e., z=1`

Therefore, `theta=1/jln[1]+c`, but `ln [1] =0` to any base, therefore `c=0`.

So we have, `theta=1/jlnz`, i.e, `jtheta=lnx`, and so `e^j theta=z` [ theory of logs], therefore from ...`[1]`, `r[cos theta+jsin theta]=re^j theta`.......`[3]`

[ Proof of De Moivres theorem, for any value of `n`]

From .....`[3]`, `cos theta+jsin theta=e^j theta` [ omitting `r`].......`[4]`

Raising both sides of ....`[3]` to the power `n`, `[cos theta+jsin theta]^n`=`e^[n j theta]`

=`e^[j[n theta]]` , but `e^[j[n theta]]`= `cosn theta+ jsin n theta` , from .......`[4]`, replacing `theta` by `n theta`,

Therefore , `[cos theta+jsin theta]^n`= `cosn theta+jsin ntheta`.